[LeetCode 322] Coin Change

最新推荐文章于 2022-07-14 16:42:52 发布

aique3350

最新推荐文章于 2022-07-14 16:42:52 发布

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原文链接：http://www.cnblogs.com/lz87/p/7006859.html

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本文详细解析了硬币找零问题，通过动态规划的方法找到组成特定金额所需的最少硬币数量。介绍了状态转移方程及代码实现，并对比了背包VI和组合总和IV等问题。

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You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

Example 1:
coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)

Example 2:
coins = [2], amount = 3
return -1.

Note:
You may assume that you have an infinite number of each kind of coin.

BackPack VI and Combination Sum IV: Find all possible combinations that sum to a target value

Coin Change: Find the combination that sums to a target value and uses the fewest number of elements

State: dp[i]: the fewest number of coins needed that sum to i

Function: dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1), if i >= coins[j] && dp[i - coins[j]] < Integer.MAX_VALUE

i >= coins[j]: only consider picking a coin if its value is not greater than the target value i;

dp[i - coins[j]] < Integer.MAX_VALUE: if we did pick coins[j], then we must be able to find a combination that sums

to i - coins[j];

Initialization: dp[0] = 0, dp[i] = Integer.MAX_VALUE, for i >= 1

Answer: dp[amount] or -1

 1 public class Solution {
 2     public int coinChange(int[] coins, int amount) {
 3         if(amount <= 0){
 4             return 0;
 5         }
 6         if(coins == null || coins.length == 0){
 7             return -1;
 8         }
 9         int[] dp = new int[amount + 1];
10         dp[0] = 0;
11         for(int i = 1; i <= amount; i++){
12             dp[i] = Integer.MAX_VALUE;
13         }
14         for(int i = 1; i <= amount; i++){
15             for(int j = 0; j < coins.length; j++){
16                 if(i >= coins[j] && dp[i - coins[j]] < Integer.MAX_VALUE){
17                     dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
18                 }
19             }
20         }
21         if(dp[amount] < Integer.MAX_VALUE){
22             return dp[amount];
23         }
24         return -1;
25     }
26 }